Question:
In a biprism experiment a steady interference pattern is observed on the screen using a light of wavelength \(5000 \mathop {\rm{A}}\limits^{\rm{o}}\). Without disturbing the set up of the experiment, the source of light is replaced by a source of wavelength \(6400 \mathop {\rm{A}}\limits^{\rm{o}}\).
The fringe width will
In a biprism experiment a steady interference pattern is observed on the screen using a light of wavelength \(5000 \mathop {\rm{A}}\limits^{\rm{o}}\). Without disturbing the set up of the experiment, the source of light is replaced by a source of wavelength \(6400 \mathop {\rm{A}}\limits^{\rm{o}}\).
The fringe width will
Show Hint
Fringe width is directly proportional to the wavelength when everything else is unchanged.
Updated On: Nov 13, 2026
- decrease by \(48\%\)
- decrease by \(28\%\)
- increase by \(48\%\)
- increase by \(28\%\)
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept
In a biprism (or double slit) setup the fringe width is \(\beta=\dfrac{\lambda D}{d}\). With the setup unchanged, \(\beta\propto\lambda\).
Step 2: Take the ratio
\(\dfrac{\beta_2}{\beta_1}=\dfrac{6400}{5000}=1.28\)
Step 3: Calculation
The fringe width becomes 1.28 times, an increase of \(0.28=28\%\). This matches option (D).
Final Answer:
The fringe width increases by 28 percent. \[ \boxed{\text{(D) }\text{increase by }28\%} \]
In a biprism (or double slit) setup the fringe width is \(\beta=\dfrac{\lambda D}{d}\). With the setup unchanged, \(\beta\propto\lambda\).
Step 2: Take the ratio
\(\dfrac{\beta_2}{\beta_1}=\dfrac{6400}{5000}=1.28\)
Step 3: Calculation
The fringe width becomes 1.28 times, an increase of \(0.28=28\%\). This matches option (D).
Final Answer:
The fringe width increases by 28 percent. \[ \boxed{\text{(D) }\text{increase by }28\%} \]
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